Anna, the older twin, was
born three hours before her younger sister Lana. On Lana's 21st
birthday,

they both went out to a club
with some friends. The bouncer was checking ID's and let Lana
go in.

But when the older Anna tried
to follow her in, the bouncer said, "I'm sorry, your 21st
birthday

isn't for another two days."How is this possible? Explain
fully and correctly to win the contest.

Solution: (by
Andy and Aaron Murdock)

For this problem, the "impossible
twins" aren't impossible just improbable. They were both
given birth on a boat (or a plane or helicopter, etc, just anything
capable of crossing the international dateline). Anna, the older
twin, was born on March 1 of a non-leap-year on the Asian side
of the International Dateline, three hours later, after they
had crossed to the other side of the Dateline, Lana was born,
the date being February 28. 21 years later, which happens to
be a leap-year, they go to a night club on the 28th of February,
Lana's birthday, yet Anna has to wait two days, the 29th and
then the 1st.

The student parking lot has
81 cars in it; all Acuras, Beetles, and Camrys. There are half
as many Acuras as Beetles, and the number of Camrys is 80% of
the number of Acuras and Beetles together. How many of each kind
of car is in the parking lot? Show
all steps to win contest!

Solution: (by Andy M & George C)

Let A = # of Acuras, B = # of Beetles, and C = # of Camrys.

Then A + B + C = 81 , B = 2 A , C = .8 (A + B).

Subbing in gives A + 2 A + .8 (A + 2 A) = 81 , 5.4 A = 81
,

A = 15 , B = 2 (15) = 30 , C = .8 (45) = 36

Answer: 15 Acuras, 30 Beetles, and 36 Camrys.

We now have respondents from as far away as my old stomping
grounds of Columbus, Ohio. Welcome!

(I had two requests to make
the problems more challenging. So try this week's problem and
advertise it to your smart friends!)

A spider, in the top-left-front
corner of a 10 x 10 x 10 foot room, sees a big fat fly in the
bottom-right-back corner. Describe the shortest path, and the
length of the path, that the spider can crawl to get the fly.That's crawl, not jump, fly or spider-web
express! Your explanation must be clear. (Not affiliated with
the squished fly from Problem #2.) ;-}

Solution: Think of unfolding the room and seeing the
ceiling and right wall as a 10 x 20 domino. The shortest path
is a straight line; the diagonal of this domino, which is -/(10^2
+ 20^2) = -/500 = 10-/5 ft , or about 22.36 feet.

Answer: 10-/5 ft , which is 22.36 ft to nearest hundredth.

Flavio's answer:

Note: This
is a simplified version of a famous problem, in which the room
is 12 x 12 x 30 ft, the spider is 1 ft below the ceiling in the
middle of one 12 x 12 ft end wall, and the fly is 1 ft above
the floor in the middle of the opposite end wall. Now what's
the shortest path? There are different ways the story can "unfold"
here!

A basketball playoff game
started between 3pm and 4pm, and ended between 6pm and 7pm.

The positions of the minute
hand and the hour hand were reversed at the end of the game,

compared to the beginning.
What was the exact time the game started and ended,

and how long was the game?
(Try to give exact times, not rounded
to the nearest anything.)

Solution: (by Dan
Bach; no contestants invoked the power of algebra!)

Ok, we know the game started a little after 3:30, since the
minute hand was between 6 and 7. Let's make up a variable: let
"t" be the number of minutes after 3:00 ; we figure
30 < t < 35. One minute of time is 360/60 = 6 degrees of
angle for the minute hand, and 1/12 of that, 1/2 degree, for
the hour hand. So at the start, while the minute hand is at 6
t degrees, the hour hand starts at 3:00 (90 degrees) and after
t min is at 90 + t/2 degrees.

Now at the end of the game, a little after 6:15, the minute
hand must be at 90 + t/2 degrees, which is

(90 n+ t/2) / 6 = 15 + t/12 min past 6, causing the hour
hand to move

(90 + t/2) / 12 = 15/2 + t/24 degrees past 6, which has to
match up with the original 6 t degrees:

In this year's Coupe du Monde
98, there are 4 teams in each Group, and they each play each
of the other 3 teams once. Here are the final "Pts standings"
of Groupe C, with the W . L . T
. PTS records (a win is 3 pts and a tie is 1 pt):

France. . . . . . . 3 . . 0
. . 0 . . 9

Denmark . . . . 1 . . 1 . .
1 . . 4

South Africa . .0 . . 1 . .
2 . . 2

Saudi Arabia. . 0 . . 2 . .
1 . . 1

How many "Pts standings"
are possible, and are any gettable in more than one way? (not counting order or particular teams) This one would be called 9-4-2-1.

Partial Solution: By Andy Murdock -

For the number of total possible
combinations, I got 29 -

If no teams tie, then all
scores have to be multiples of three with a sum of 18 total:

9-6-3-0, 9-3-3-3, 6-6-6-0,
6-6-3-3

If there is only one tie,
the sum is now 17 and each must have 2 occurrences

of 1,4,or 7 (each being a
1+a multiple of 3):

9-6-1-1, 9-4-4-0, 9-4-3-1, 7-7-3-0, 7-6-4-0, 7-6-3-1, 7-4-3-3

With two ties, the sum is
now 16, and similar to the last case where there

were 2 cases of one being
added to a multiple of three, it happens four times here: